final chem reviewer

8/9/2019 Final Chem Reviewer

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Sophomored 0910 Chem

I. Mole Concept- a counting unit used in chemistry to measure the ammount of substance- abbreviated as mol- Founded on Avogadro's Number

- 6.02 X 10^(23)- founder: Italian physicist Amedeo Avogadro- is equivalent to the number of atoms in a carbon-12 atom

- Different masses are used

1. Formula Mass:- a measurement of an object's mass relative to that of a carbon-12- measured in amu- used for ionic compounds.

- Ionic compounds are METAL NONMETAL1. Molecular Mass

- a measurement of an objects mass relative to that of a carbon 12 isotope- measured in amu- used for covalent compunds

- NONMETAL NONMETAL2. Molar Mass

- mass in grams numerically equal to formula or molecular mass- Measured in grams per mol, g/mol

- SOLVING FOR ATOMIC MASS (Given just the compound)1. Example: Al2 (SO4)3

Element Number of Molecules Atomic Mass (NoM X Atomic Mass)

Aluminum 2 26.98 53.96

Sulfer 3 32.07 96.21

Oxygen 12* 16 192

Total: 342.17 amu

*: its twelve because 4 times 3 is 12, 4 being the subscript of O and 3 being the subscript of the sulfur AND oxygen

1. Example: SF6

Element Number of Molecules Atomic Mass (NoM X Atomic Mass)

Sulfur 1 32.07 32.07

Fluorine 6 19 114Total: 146.07 amu

1.SOLVING FOR MOLE, PARTICLES/ATOMS, GRAMS1. Note: Use this naming system1 element = atoms (8 atoms of oxygen)

compounds = particles (8 particles of Freon)molecular mass = molecules (8 molecules of H2O)formula mass = formula units A.K.A fu (8 fu CaCl)

1.Conversion Chart

----- ( P/AN)---> ---- (mol X amu) ---->Particle Mole Grams


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1. in other words..... the percentage of an element in a compound- HOW TO SOLVE FOR PERCENT COMPOSITION

2. Example: (NH4)2 SO4

Element Number ofMolecules

Atomic Mass (NoM X Atomic Mass) (


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- the wheels are excess, because they are left over after thereaction has completed- the frame is the limiter, because it determines how much product is produced

- Scarier example- 4 (NH3) + 5 (O2) ? 4 (NO) + 6 (H2O)

- Assume that 4.50g of NH3 reacts with 7.50g Oxygen- What is the limiter?

- Step 1: Solve for the product- here, you will solve for the product two times, 1 for each of the reactants. Stoich will be used

(NH3)

14.50g NH3 ---(convert to mole)--> X 6 H2O --(mass)--> = 7.124g H2O4 NH3

(O2)(same process as above, just replace 4.50g NH3 with 7.50g O2) = 5.07g H2

Note that the product of NH3 is greater than O2. That makes it excess, and that being the case, O2 is thelimiter

- When asked for the amount of product used, always use the limiting reactant's assumed product. In the previous example, wewill be using 5.07g H2O as the estimated product

- When asked for how much of the excess reactant was used....- Perform another stoich equation, but with the excess reactant instead of the product

1. 7.50g O2 (mole)--> X 4 NH3 ---(mass)---> = 3.19g (7.16g/6.97) X 100 = 97.35% Yield

- an even harder example1. Wine is produced by fermentation

C6 H12 O6 2(C2 H6 O) + 2 (CO2)

- given 983g fructose (C6 H12 O6) what is the percent yield if 327g of ethanol was used?

- HOW TO SOLVE FOR THE YIELDS:Step 1: BalanceStep 2: Stoich. This equation will convert ethanol to fructose, or the given reactant to the unknown product=> 938g F X 1mol X 2 E X 46.07g = 480g E ; (327/480) X 100= 97.35%

180.16g 1 F 1 molLegend: Fructose,Mole Conversion, Ethanol, Mass Conversion, Theoretical Yield Actual, Percent

Answers:Theo: 480Percent: 97.35

VIII. Acids and Bases


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ACIDS BASEpH Below 7 Above 7

pOH Above 7 Below 7

According to

Arthenius Contains H+ Contains OH-

Bronsted-Laury Proton Donor Proton Acceptor

Lewis Electron Acceptor Electron Doner

Concentration

1. the amount of H or OH in an acid/base2. measured in M (Molarity)3. Use these formulas

1. H+ to1. pH: -log(H)

2. OH- to1. pOH: -log(OH-)

3. pH to1. H+: 10^(-pH)2. pOH: 14 pH

4. pOH to1. OH-: 10^(-pOH)2. pH: 14- pOH

Note: Direct conversion from H to OH and vice versa is IMMPOSSIBLE4. Examples

1. Given: pH = 12; unknown H

=> 10^(-3.12)=> 7.58 X 10^(-4) = H

2. Given: H = 0.15M; unknown pH=> -log(0.15)=> 0.82 pH

IX. Solutionshomogenous mixture of two or more substances

1. homogenous: parts cannot be defined. It will not separate if left alone2 components

1. Solute: the dissolved substance2. Solvent: the dissolving substance

Solvation: the reaction between a solute and solvent2 Main Branches

1. Physical State1. Gaseous

1. gas in gas

2. Liquid1. liquid in liquid2. gas in liquid3. solid in liquid

3. Solid1. solid in solid2. gas in solid3. liquid in solid

2. Solute Concentration1. Saturated: a solution containing the maximum amount of solute2. Unsaturated: a solution containing less than the maximum amount of solute3. Supersaturated: a solution containing more than the maximum amount of solute

Solubility1. Refers to the maximum amount of solute expressed in grams that can be dissolved in 100g of water

1. soluble: the substance can be dissolved2. insoluble: the substance cannot be dissolved

2. Solubility Rules1. All salts of alkali metals (Group 1) and ammonium ion NH4+ are soluble2. All salts containing acetates, per chlorates, chlorate, and nitrate are soluble3. Salts of silver, lead, and mercury (I) are insoluble unless affected by Rule #24. Chlorides, bromides, and iodides are soluble unless affected by Rule #35. Carbonates. Hydroxides, oxides, phosphates, silicates, and sulfides of metals are insoluble unless affected by

Rule #16. Sulfates are soluble except for sulfates of calcium and barium

Miscibility1. refers to the ability of a liquid to dissolve another liqud

1. miscible: the liquid can be dissolved2. immiscible: the liquid cannot be dissolved


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